Intergrate 1/√(9-x^2)?

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Answer 1 · 2018-04-29T19:19:54

#I=sin^-1(x/3)+c#

Here,

#I=int1/sqrt(9-x^2)dx=int1/sqrt(3^2-x^2)dx#

Use the formula:

#color(red)(int1/sqrt(a^2-x^2)dx=sin^-1(x/a)+c),where, a=3#

#:.I=sin^-1(x/3)+c#

If you are not familiar with above formula,then use the method given below.

Let, #x=3sinu=>dx=3cosu#

#and sinu=x/3=>u=sin^-1(x/3)#

#:.I=int1/sqrt(3^2-3^2sin^2x)3cosudu#

#=int(3cosu)/(3sqrt(1-sin^2u))du#

#=int(3cosu)/(3cosu)du#

#=int1du#

#I=u+c#

Subst. back , #u=sin^-1(x/3)#

#I=sin^-1(x/3)+c#